Write all the other trigonometric ratios of \(\angle A\) in terms of sec A.
\( \sin A = \dfrac{\sqrt{\sec^2 A - 1}}{\sec A} \)
\( \cos A = \dfrac{1}{\sec A} \)
\( \tan A = \sqrt{\sec^2 A - 1} \)
\( \cot A = \dfrac{1}{\sqrt{\sec^2 A - 1}} \)
\( \text{cosec } A = \dfrac{\sec A}{\sqrt{\sec^2 A - 1}} \)
Goal: Express all other trigonometric ratios of \(\angle A\) in terms of \(\sec A\).
Step 1: Start with the definition of secant
We know:
\(\sec A = \dfrac{1}{\cos A}\)
So, directly:
\(\cos A = \dfrac{1}{\sec A}\)
Step 2: Use the identity connecting \(\sec\) and \(\tan\)
One very important identity is:
\(\sec^2 A = 1 + \tan^2 A\)
Rearrange it to get \(\tan^2 A\):
\(\tan^2 A = \sec^2 A - 1\)
Now take square root on both sides:
\(\tan A = \sqrt{\sec^2 A - 1}\)
Student Note: In school problems like this, we usually take the positive root because \(A\) is taken as an acute angle in a right triangle (so trigonometric ratios are positive).
Step 3: Find \(\sin A\) using \(\tan A\) and \(\sec A\)
We know:
\(\tan A = \dfrac{\sin A}{\cos A}\) and \(\sec A = \dfrac{1}{\cos A}\).
Divide \(\tan A\) by \(\sec A\):
\(\dfrac{\tan A}{\sec A} = \dfrac{\sin A/\cos A}{1/\cos A} = \sin A\)
So,
\(\sin A = \dfrac{\tan A}{\sec A}\)
Substitute \(\tan A = \sqrt{\sec^2 A - 1}\):
\(\sin A = \dfrac{\sqrt{\sec^2 A - 1}}{\sec A}\)
Step 4: Find \(\cot A\) and \(\text{cosec } A\) using reciprocals
\(\cot A\) is the reciprocal of \(\tan A\):
\(\cot A = \dfrac{1}{\tan A} = \dfrac{1}{\sqrt{\sec^2 A - 1}}\)
\(\text{cosec } A\) is the reciprocal of \(\sin A\):
\(\text{cosec } A = \dfrac{1}{\sin A} = \dfrac{1}{\dfrac{\sqrt{\sec^2 A - 1}}{\sec A}} = \dfrac{\sec A}{\sqrt{\sec^2 A - 1}}\)
Final Results (in terms of \(\sec A\)):
\(\cos A = \dfrac{1}{\sec A}\)
\(\tan A = \sqrt{\sec^2 A - 1}\)
\(\sin A = \dfrac{\sqrt{\sec^2 A - 1}}{\sec A}\)
\(\cot A = \dfrac{1}{\sqrt{\sec^2 A - 1}}\)
\(\text{cosec } A = \dfrac{\sec A}{\sqrt{\sec^2 A - 1}}\)